all chapter Statistics for Engineers and Scientists William Navidi 5th edition solutions manual pdf PDF

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Authors: William Navidi
Published: McGraw-Hill 2019
Edition: 5th
Pages: 502
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FOLFNKHUHWRGRZQORDG

Solutions Manual to accompany

STATISTICS FOR ENGINEERS AND SCIENTISTS, 5th ed. Prepared by

William Navidi

PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and protected by copyright and other state and federal laws. By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw-Hill: This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook. No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or used by any student or other third party. No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of McGraw-Hill.

PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

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FOLFNKHUHWRGRZQORDG

Table of Contents

Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239 Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278 Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355 Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405 Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 441 Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483

PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

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SECTION 1.1

FOLFNKHUHWRGRZQORDG

1

Chapter 1 Section 1.1 1. (a) The population consists of all the times the process could be run. It is conceptual. (b) The population consist of all the registered voters in the state. It is tangible. (c) The population consist of all people with high cholesterol levels. It is tangible. (d) The population consist of all concrete specimens that could be made from the new formulation. It is conceptual. (e) The population consist of all bolts manufactured that day. It is tangible.

2.

(iii). It is very unlikely that students whose names happen to fall at the top of a page in the phone book will differ systematically in height from the population of students as a whole. It is somewhat more likely that engineering majors will differ, and very likely that students involved with basketball intramurals will differ.

3. (a) False (b) True

4. (a) False (b) True

5. (a) No. What is important is the population proportion of defectives; the sample proportion is only an approximation. The population proportion for the new process may in fact be greater or less than that of the old process. (b) No. The population proportion for the new process may be 0.12 or more, even though the sample proportion was only 0.11. (c) Finding 2 defective circuits in the sample.

6. (a) False

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2

CHAPTER 1

(b) True (c) True

7.

A good knowledge of the process that generated the data.

8. (a) An observational study (b) It is not well-justified. Because the study is observational, there could be differences between the groups other than the level of exercise. These other differences (confounders) could cause the difference in blood pressure. 9. (a) A controlled experiment (b) It is well-justified, because it is based on a controlled experiment rather than an observational study.

Section 1.2 1.

False

2.

No. In the sample 1, 2, 4, the mean is 7/3, which does not appear at all.

3.

No. In the sample 1, 2, 4, the mean is 7/3, which does not appear at all.

4.

No. The median of the sample 1, 2, 4, 5 is 3.

5.

The sample size can be any odd number.

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FOLFNKHUHWRGRZQORDG

SECTION 1.2

6.

Yes. For example, the list 1, 2, 12 has an average of 5 and a standard deviation of 6.08.

7.

Yes. If all the numbers in the list are the same, the standard deviation will equal 0.

8.

The mean increases by $50; the standard deviation is unchanged.

9.

The mean and standard deviation both increase by 5%.

3

10. (a) Let 𝑋1 , ..., 𝑋100 denote the 100 numbers of occupants. 100 ∑

𝑋𝑖 = 70(1) + 15(2) + 10(3) + 3(4) + 2(5) = 152

𝑖=1

𝑋=

∑100 𝑖=1 𝑋𝑖 100

=

152 = 1.52 100

(b) The sample variance is ) ( 100 2 1 ∑ 2 𝑋 − 100𝑋 = 99 𝑖=1 𝑖 1 [(70)12 + (15)22 + (10)32 + (3)4 2 + (2)52 − 100(1.522 )] = 99 = 0.87838 √ The standard deviation is 𝑠 = 𝑠2 = 0.9372. Alternatively, the sample variance can be computed as 𝑠2

𝑠2

100 1 ∑ (𝑋 − 𝑋)2 99 𝑖=1 𝑖 1 = [70(1 − 1.52)2 + 15(2 − 1.52)2 + 10(3 − 1.52)2 + 3(4 − 1.52)2 + 2(5 − 1.52)2 ] 99 = 0.87838

=

(c) The sample median is the average of the 50th and 51st value when arranged in order. Both these values are equal to 1, so the median is 1.

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4

CHAPTER 1

(d) The first quartile is the average of the 25th and 26th value when arranged in order. Both these values are equal to 1, so the first quartile is 1. The third quartile is the average of the 75th and 76th value when arranged in order. Both these values are equal to 2, so the first quartile is 2.

(e) Of the 100 cars, 15 + 10 + 3 + 2 = 30 had more than the mean of 1.52 occupants, so the proportion is 30/100 = 0.3. (f) The quantity that is one standard deviation greater than the mean is 1.52 + 0.9372 = 2.5472. Of the 100 cars, 10 + 3 + 2 = 15 had more than 2.8652 children, so the proportion is 15/100 = 0.15.

(g) The region within one standard deviation of the mean is 1.52 ± 0.9372 = (0.5828, 2.4572). Of the 100 cars, 70 + 15 = 85 are in this range, so the proportion is 85/100 = 0.85.

11.

The total height of the 20 men is 20 × 178 = 3560. The total height of the 30 women is 30 × 164 = 4920. The total height of all 50 people is 3560 + 4920 = 8480. There are 20 + 30 = 50 people in total. Therefore the mean height for both groups put together is 8480∕50 = 169.6 cm.

12. (a) The mean for A is (18.0+18.0+18.0+20.0+22.0+22.0+22.5+23.0+24.0+24.0+25.0+25.0+25.0+25.0+26.0+26.4)∕16 = 22.744 The mean for B is (18.8+18.9+18.9+19.6+20.1+20.4+20.4+20.4+20.4+20.5+21.2+22.0+22.0+22.0+22.0+23.6)∕16 = 20.700 The mean for C is (20.2+20.5+20.5+20.7+20.8+20.9+21.0+21.0+21.0+21.0+21.0+21.5+21.5+21.5+21.5+21.6)∕16 = 20.013 The mean for D is (20.0+20.0+20.0+20.0+20.2+20.5+20.5+20.7+20.7+20.7+21.0+21.1+21.5+21.6+22.1+22.3)∕16 = 20.806

(b) The median for A is (23.0 + 24.0)/2 = 23.5. The median for B is (20.4 + 20.4)/2 = 20.4. The median for C is (21.0 + 21.0)/2 = 21.0. The median for D is (20.7 + 20.7)/2 = 20.7.

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FOLFNKHUHWRGRZQORDG

SECTION 1.2

5

(c) 0.20(16) = 3.2 ≈ 3. Trim the 3 highest and 3 lowest observations. The 20% trimmed mean for A is (20.0 + 22.0 + 22.0 + 22.5 + 23.0 + 24.0 + 24.0 + 25.0 + 25.0 + 25.0)∕10 = 23.25 The 20% trimmed mean for B is (19.6 + 20.1 + 20.4 + 20.4 + 20.4 + 20.4 + 20.5 + 21.2 + 22.0 + 22.0)∕10 = 20.70 The 20% trimmed mean for C is (20.7 + 20.8 + 20.9 + 21.0 + 21.0 + 21.0 + 21.0 + 21.0 + 21.5 + 21.5)∕10 = 21.04 The 20% trimmed mean for D is (20.0 + 20.2 + 20.5 + 20.5 + 20.7 + 20.7 + 20.7 + 21.0 + 21.1 + 21.5)∕10 = 20.69

(d) 0.25(17) = 4.25. Therefore the first quartile is the average of the numbers in positions 4 and 5. 0.75(17) = 12.75. Therefore the third quartile is the average of the numbers in positions 12 and 13. A: 𝑄1 = 21.0, 𝑄3 = 25.0; B: 𝑄1 = 19.85, 𝑄3 = 22.0; C: 𝑄1 = 20.75, 𝑄3 = 21.5; D: 𝑄1 = 20.1, 𝑄3 = 21.3 (e) The variance for A is 1 [18.02 + 18.02 + 18.02 + 20.02 + 22.02 + 22.02 + 22.52 + 23.02 + 24.02 15 + 24.02 + 25.02 + 25.02 + 25.02 + 25.02 + 26.02 + 26.42 − 16(22.7442 )] = 8.2506 √ The standard deviation for A is 𝑠 = 8.2506 = 2.8724. The variance for B is 𝑠2

=

1 [18.82 + 18.92 + 18.92 + 19.62 + 20.12 + 20.42 + 20.42 + 20.42 + 20.42 15 + 20.52 + 21.22 + 22.02 + 22.02 + 22.02 + 22.02 + 23.62 − 16(20.7002 )] = 1.8320 √ The standard deviation for B is 𝑠 = 1.8320 = 1.3535. The variance for C is 1 [20.22 + 20.52 + 20.52 + 20.72 + 20.82 + 20.92 + 21.02 + 21.02 + 21.02 𝑠2 = 15 + 21.02 + 21.02 + 21.52 + 21.52 + 21.52 + 21.52 + 21.62 − 16(20.0132 )] = 0.17583 √ The standard deviation for C is 𝑠 = 0.17583 = 0.4193. The variance for D is 1 𝑠2 = [20.02 + 20.02 + 20.02 + 20.02 + 20.22 + 20.52 + 20.52 + 20.72 + 20.72 15 + 20.72 + 21.02 + 21.12 + 21.52 + 21.62 + 22.12 + 22.32 − 16(20.8062 )] = 0.55529 √ The standard deviation for D is 𝑠 = 0.55529 = 0.7542. 𝑠2

=

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6

CHAPTER 1

(f) Method A has the largest standard deviation. This could be expected, because of the four methods, this is the crudest. Therefore we could expect to see more variation in the way in which this method is carried out, resulting in more spread in the results.

(g) Other things being equal, a smaller standard deviation is better. With any measurement method, the result is somewhat different each time a measurement is made. When the standard deviation is small, a single measurement is more valuable, since we know that subsequent measurements would probably not be much different.

13. (a) All would be divided by 2.54.

(b) Not exactly the same, because the measurements would be a little different the second time.

14. (a) We will work in units of $1000. Let 𝑆0 be the sum of the original 10 numbers and let 𝑆1 be the sum after the change. Then 𝑆0 ∕10 = 70, so 𝑆0 = 700. Now 𝑆1 = 𝑆0 −100 +1000 = 1600, so the new mean is 𝑆1 ∕10 = 160. (b) The median is unchanged at 55. ∑10 2 𝑋𝑖 . Then the variance is (1∕9)[𝑇0 − 10(702 )] = (c) Let 𝑋1 , ..., 𝑋10 be the original 10 numbers. Let 𝑇0 = 𝑖=1 2 2 2 20 = 400, so 𝑇0 = 52,600. Let 𝑇1 be the sum √ of the squares after the change. Then 𝑇1 = 𝑇0 − 100 + 1000 = 2 1, 042, 600. The new standard deviation is (1∕9)[𝑇1 − 10(160 )] = 295.63.

15. (a) The sample size is 𝑛 = 16. The tertiles have cutpoints (1∕3)(17) = 5.67 and (2∕3)(17) = 11.33. The first tertile is therefore the average of the sample values in positions 5 and 6, which is (44 + 46)∕2 = 45. The second tertile is the average of the sample values in positions 11 and 12, which is (76 + 79)∕2 = 77.5.

(b) The sample size is 𝑛 = 16. The quintiles have cutpoints (𝑖∕5)(17) for 𝑖 = 1, 2, 3, 4. The quintiles are therefore the averages of the sample values in positions 3 and 4, in positions 6 and 7, in positions 10 and 11, and in positions 13 and 14. The quintiles are therefore (23 + 41)∕2 = 32, (46 + 49)∕2 = 47.5, (74 + 76)∕2 = 75, and (82 + 89)∕2 = 85.5.

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FOLFNKHUHWRGRZQORDG

SECTION 1.3

7

16. (a) Seems certain to be an error. (b) Could be correct.

Section 1.3 Stem 0 1 1. (a) 2 3 4 5 6 7 8 9 10 11 12 13

Leaf 011112235677 235579 468 11257 14699 5 16 9 0099

0 7 7 0.45 0.4

Relative Frequency

0.35

(b) Here is one histogram. Other choices for the endpoints are possible.

0.3 0.25 0.2 0.15 0.1 0.05 0 0

(c)

0

2

4

6 8 Rainfall (inches)

10

2

4

12

6 8 10 Rainfall (inches)

12

14

14

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FOLFNKHUHWRGRZQORDG

8 (d)

CHAPTER 1

Rainfall (inches)

15

10

The boxplot shows one outlier. 5

0

2. (a) Stem 14 15 16 17 18 19 20 21 22

Leaf 12333444467788889 0023344567799 124456 2222338

9 4

(b) Here is one histogram. Other choices for the endpoints are possible.

0.35

Relat ive Frequency

0.3 0.25 0.2 0.15 0.1 0.05 0 14

15

16

17 18 19 20 21 Sulfur Trioxide (percent)

22

23

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SECTION 1.3

(c)

(d)

14

16

18

20

22

9

24

24 22 20

The boxplot shows 2 outliers. 18 16 14

3.

Stem 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

Leaf 1588 00003468 0234588 0346 2235666689 00233459 113558 568 1225 1 2 06

1 6 9

3

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10

CHAPTER 1

There are 23 stems in this plot. An advantage of this plot over the one in Figure 1.6 is that the values are given to the tenths digit instead of to the ones digit. A disadvantage is that there are too many stems, and many of them are empty.

4. (a) Here are histograms for each group. Other choices for the endpoints are possible. 0.6

Relative Frequency

Relative Frequency

0.4 0.3 0.2 0.1

0.5 0.4 0.3 0.2 0.1

0

(b)

50

0

150 250 350 450 550 Concentration (mg/kg)

50 150 250350450550650750850950

Concentration (mg/kg)

1000 800 600 400 200 0 Chromium

Nickel

(c) The concentrations of nickel are on the whole lower than the concentrations of chromium. The nickel concentrations are highly skewed to the right, which can be seen from the median being much closer to the first quartile than the third. The chromium concentrations are somewhat less skewed. Finally, the nickel concentrations include an outlier.

Page 10 PROPRIETARY MATERIAL. © T...